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NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

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Ex 6.5 Class 7 Maths Question 1.
PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 1
Solution:
In right angled triangle PQR, we have
QR2 = PQ2 + PR2 From Pythagoras property)
= (10)2 + (24)2
= 100 + 576 = 676
∴ QR = \(\sqrt{676}\) = 26 cm
The, the required length of QR = 26 cm.

Ex 6.5 Class 7 Maths Question 2.
ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 2
Solution:
In right angled∆ABC, we have
BC2 + (7)2 = (25)2 (By Pythagoras property)
⇒ BC2 + 49 = 625
⇒ BC2 = 625 – 49
⇒ BC2 = 576
∴ BC = \(\sqrt{576}\) = 24 cm
Thus, the required length of BC = 24 cm.

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    Ex 6.5 Class 7 Maths Question 3.
    A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 3
    Solution:
    Here, the ladder forms a right angled triangle.
    ∴ a2 + (12)2 = (15)2 (By Pythagoras property)
    ⇒ a2+ 144 = 225
    ⇒ a2 = 225 – 144
    ⇒ a2 = 81
    ∴ a = \(\sqrt{81}\) = 9 m
    Thus, the distance of the foot from the ladder = 9m

    Ex 6.5 Class 7 Maths Question 4.
    Which of the following can be the sides of a right triangle?
    (i) 2.5 cm, 6.5 cm, 6 cm.
    (ii) 2 cm, 2 cm, 5 cm.
    (iii) 1.5 cm, 2 cm, 2.5 cm
    Solution:
    (i) Given sides are 2.5 cm, 6.5 cm, 6 cm.
    Square of the longer side = (6.5)2 = 42.25 cm.
    Sum of the square of other two sides
    = (2.5)2 + (6)2 = 6.25 + 36
    = 42.25 cm.
    Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.
    ∴ The given sides form a right triangle.

    (ii) Given sides are 2 cm, 2 cm, 5 cm .
    Square of the longer side = (5)2 = 25 cm Sum of the square of other two sides
    = (2)2 + (2)2 =4 + 4 = 8 cm
    Since 25 cm ≠ 8 cm
    ∴ The given sides do not form a right triangle.

    (iii) Given sides are 1.5 cm, 2 cm, 2.5 cm
    Square of the longer side = (2.5)2 = 6.25 cm Sum of the square of other two sides
    = (1.5)2 + (2)2 = 2.25 + 4
    Since 6.25 cm = 6.25 cm = 6.25 cm
    Since the square of longer side in a triangle is equal to the sum of square of other two sides.
    ∴ The given sides form a right triangle.

    Ex 6.5 Class 7 Maths Question 5.
    A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree . Find the original height of the tree.
    Solution:
    Let AB be the original height of the tree and broken at C touching the ground at D such that
    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 4
    AC = 5 m
    and AD = 12 m
    In right triangle ∆CAD,
    AD2 + AC2 = CD2 (By Pythagoras property)
    ⇒ (12)2 + (5)2 = CD2
    ⇒ 144 + 25 = CD2
    ⇒ 169 = CD2
    ∴ CD = \(\sqrt{169}\) = 13 m
    But CD = BC
    AC + CB = AB
    5 m + 13 m = AB
    ∴ AB = 18 m .
    Thus, the original height of the tree = 18 m.

    Ex 6.5 Class 7 Maths Question 6.
    Angles Q and R of a APQR are 25° and 65°. Write which of the following is true.
    (i) PQ2 + QR2 = RP2
    (ii) PQ2 + RP2 = QR2
    (iii) RP2 + QR2 = PQ2
    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 5
    Solution:
    We know that
    ∠P + ∠Q + ∠R = 180° (Angle sum property)
    ∠P + 25° + 65° = 180°
    ∠P + 90° = 180°
    ∠P = 180° – 90° – 90°
    ∆PQR is a right triangle, right angled at P
    (i) Not True
    ∴ PQ2 + QR2 ≠ RP2 (By Pythagoras property)
    (ii) True
    ∴ PQ2 + RP2 = QP2 (By Pythagoras property)
    (iii) Not True
    ∴ RP2 + QR2 ≠ PQ2 (By Pythagoras property)

    Ex 6.5 Class 7 Maths Question 7.
    Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 6
    Solution:
    Given: Length AB = 40 cm
    Diagonal AC = 41 cm
    In right triangle ABC, we have
    AB2 + BC2= AC2 (By Pythagoras property)
    ⇒ (40)2 + BC2 = (41)2
    ⇒ 1600 + BC2 = 1681
    ⇒ BC2 = 1681 – 1600
    ⇒ BC2 = 81
    ∴ BC = \(\sqrt{81}\) = 9 cm
    ∴ AB = DC = 40 cm and BC = AD = 9 cm (Property of rectangle)
    ∴ The required perimeter
    = AB + BC + CD + DA
    = (40 + 9 + 40 + 9) cm
    = 98 cm

    Ex 6.5 Class 7 Maths Question 8.
    The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 7
    Solution:
    Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm
    Since, the diagonals of a rhombus bisect each other at 90°.
    ∴ OA = OC = 8 cm and OB = OD = 15 cm
    In right ∆OAB,
    AB2 = OA2 + OB2 (By Pythagoras property)
    = (8)2+ (15)2 = 64 + 225
    = 289
    ∴ AB =\(\sqrt{289}\)= 17 cm
    Since AB = BC = CD = DA (Property of rhombus)
    ∴ Required perimeter of rhombus
    = 4 × side = 4 × 17 = 68 cm.

    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 Q1

    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 Q2

    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 Q3

    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 Q4

    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 Q5

    NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 Q6

     

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